To find the zeros of the polynomial function given by f(x) = 5x^2 – 82x + 83, we start by setting the function equal to zero:
5x^2 – 82x + 83 = 0
Next, we can apply the quadratic formula, which is:
x = (-b ± √(b² – 4ac)) / 2a
In our case, the coefficients are:
- a = 5
- b = -82
- c = 83
Now, we calculate the discriminant (b² – 4ac):
Discriminant = (-82)² – 4(5)(83) = 6724 – 1660 = 5064
Since the discriminant is positive, we can expect two distinct real roots. Now, let’s calculate the roots:
x = (82 ± √5064) / (2 * 5)
Now we find the square root of 5064:
√5064 ≈ 71.1 (approximately)
Plugging it back into the formula gives us:
x₁ = (82 + 71.1) / 10 ≈ 15.31
x₂ = (82 – 71.1) / 10 ≈ 1.09
Thus, the zeros of the polynomial are approximately:
- x ≈ 15.31 with multiplicity 1
- x ≈ 1.09 with multiplicity 1
In conclusion, the polynomial function f(x) = 5x^2 – 82x + 83 has two zeros: approximately 15.31 and 1.09, both with a multiplicity of 1, indicating they are simple roots.